Error Solutions Workbook
Physics

• The field B is uniform (same value everywhere in space).
• Four different actions A, B, C, D are performed on a coil sitting in that field.

Faraday's law: |ε| = dΦ/dt, and flux is Φ = B·A·cosθ. So an emf needs one of these three numbers to change: the field B, the area A, or the angle θ. Speed by itself is irrelevant. If the coil simply slides around inside a uniform field, every point of the coil always sees the same B, the area is the same, the tilt is the same — so Φ is frozen and ε = 0, no matter how fast or how jerkily it moves.

1. A — uniform speed, uniform field. B same, A same, θ same ⇒ Φ constant ⇒ ε = 0.
2. B — non-uniform speed, uniform field. Acceleration changes nothing in the flux formula. B, A, θ all still constant ⇒ ε = 0.
3. C — rotating about a diameter. Now θ = ωt, so Φ = BA cosωt ⇒ ε = BAω sinωt ⇒ emf exists. (This is exactly how an AC generator works.)
4. D — area changing. Φ = B·A(t) ⇒ ε = B·dA/dt ≠ 0 ⇒ emf exists.

• The uniform-field filter: the moment you read the words “uniform magnetic field”, cross out every option whose only feature is translation. Sliding inside a uniform field never induces emf.
• Only three verbs create emf: rotate, stretch/shrink, or change B. Translation is not on that list.
• Both distractors here (A and B) are the “motion = emf” reflex. Motion is neither necessary nor sufficient.

• φ(t) = t³ − 9t² + 24t − 30 mWb, t in seconds.
• Units (mWb) are irrelevant here — we only need where the derivative vanishes.

ε = −dφ/dt. So “emf is zero” simply means “the flux graph is momentarily flat” — i.e. at its turning points (maximum or minimum). A cubic has at most two turning points, so expect two answers, and the question wants the distance between them.

1. Differentiate: dφ/dt = 3t² − 18t + 24.
2. Set to zero: 3t² − 18t + 24 = 0.
3. Divide the whole equation by 3: t² − 6t + 8 = 0.
4. Factorise: (t − 2)(t − 4) = 0 ⇒ t = 2 s and t = 4 s.
5. Interval = 4 − 2 = 2 s.

• Always divide out the common factor first. 3t² − 18t + 24 looks ugly; t² − 6t + 8 factorises on sight (sum 6, product 8 → 2 and 4).
• Sum–product shortcut for the gap: the difference of roots is √((sum)² − 4×product) = √(36 − 32) = 2 — you never have to find the roots individually.
• The constant term (−30) can be ignored the instant you see “emf”: differentiation kills it.

• R = 2 Ω
• Straight-line graph: Δφ = 8 Wb in Δt = 4 s, so the slope is constant.

A straight-line φ–t graph means a constant emf, hence a constant current. Then:

1. A — emf: ε = 8/4 = 2 V ⇒ matches s (2).
2. B — current: i = ε/R = 2/2 = 1 A ⇒ matches q (1).
3. C — charge in 3 s: q = i·t = 1 × 3 = 3 C ⇒ matches r (3).
4. D — heat in 2 s: H = i²Rt = (1)² × 2 × 2 = 4 J ⇒ matches p (4).

• Charge has a graph shortcut: q = Δφ/R. In 3 s the flux rises by 6 Wb, so q = 6/2 = 3 C — same answer, no current needed.
• Since i = 1 A exactly, heat H = i²Rt collapses to just R×t = 2×2 = 4. Look for the “i = 1” simplification before grinding.
• Anti-swap drill: write A, B, C, D down the page as four separate lines with the number beside each, then read the options. Never map straight into an option string.

• Conducting wire of length L, pivoted at the top, bob mass m.
• Angular amplitude θ; field B perpendicular to the swing plane.

Two ideas stacked together:

• Rotating rod: a rod of length L pivoted at one end and rotating with angular speed ω develops ε = ½BωL². Note the ½ — different points of the rod move at different speeds, so you use the average.
• Energy conservation: the bob is fastest at the lowest point, where v = √(2gh) with h = L(1 − cosθ).

1. Drop height: h = L(1 − cosθ). Use the identity 1 − cosθ = 2sin²(θ/2) ⇒ h = 2L sin²(θ/2).
2. Speed at the bottom: v = √(2gh) = √(4gL sin²(θ/2)) = 2 sin(θ/2)√(gL).
3. Angular speed there: ω = v/L.
4. Rod emf: ε = ½BωL² = ½B(v/L)L² = ½BvL.
5. Substitute v: ε = ½·B·L·2 sin(θ/2)√(gL) = BL sin(θ/2)√(gL).

• Dimension sieve first: emf ~ B·L·v, and v must carry √(gL). Any option with (gL)^3/2 or (gL)² is dimensionally dead — two options gone in three seconds.
• The remaining choice is only between 2BL… and BL…. That factor of 2 is precisely the ½ of the rotating rod. If you forget the rod is rotating (not sliding), you get the wrong one.

• EMF = 18 V.
• Resistors 10, 20, 30, 40, 80 Ω; inductors 10 H and 20 H.
• Instant considered: t = 0⁺, switch just closed.

An inductor hates sudden change. Current through an inductor cannot jump, so if it was zero before the switch closed, it is still exactly zero at t = 0⁺. Electrically, at t = 0 an inductor behaves like an open circuit / broken wire. (At t = ∞ it becomes a plain wire — the opposite extreme.)

1. At t = 0, replace both inductors by breaks. That kills the 10 Ω branch (blocked by the 10 H) and the 80 Ω branch (blocked by the 20 H).
2. Trace what is left as a continuous conducting path from the cell: 40 Ω → 30 Ω → 20 Ω and back to the cell.
3. All three are in series: R_eq = 40 + 30 + 20 = 90 Ω.
4. I = V/R_eq = 18/90.
5. I = 0.2 A.

• Two-extreme rule — memorise this pair and half of all L–C circuit questions become one-liners:
  t = 0 : inductor = OPEN, capacitor = SHORT t = ∞ : inductor = SHORT, capacitor = OPEN
• Reverse-engineer the arithmetic: the options are 20 A, 0.2 A, 0.5 A, 5 A → required resistances are 0.9, 90, 36, 3.6 Ω. Only 90 is buildable from 10/20/30/40/80, and 40+30+20 lands on it immediately.

• i = t + 2 ⇒ at t = 0, i = 2 A.
• di/dt = 1 A/s (constant — the current is rising).
• Cell 10 V (traversed from − to +, i.e. a potential gain of 10 V going A→B), L = 1 H, R = 3 Ω.

Walk from A to B and bookkeep every element:

• Cell: − to + ⇒ potential rises by 10 V.
• Inductor: moving along the current with current increasing ⇒ potential drops by L·di/dt.
• Resistor: moving along the current ⇒ potential drops by iR.

Then V_A + (rises) − (drops) = V_B.

1. Write the walk: V_A + 10 − L(di/dt) − iR = V_B.
2. Insert numbers at t = 0: L(di/dt) = 1 × 1 = 1 V; iR = 2 × 3 = 6 V.
3. V_A + 10 − 1 − 6 = V_B.
4. Rearrange: V_A − V_B = −10 + 1 + 6.
5. V_A − V_B = −3 V.

• Both ±3 are offered — that is the examiner telling you the magnitude is easy and the sign is the question. Slow down exactly there.
• Walk-and-tally method: draw the branch, put an arrow for i, then write a running total: 0 → +10 → +9 → +3. That final +3 is V_B − V_A, so V_A − V_B = −3. Reading the tally backwards is the single most common slip.
• Remember: a rising current makes the inductor behave like a back-emf cell opposing you — it eats potential, just like a resistor.

• Total length ℓ, axis not at an end and not at the centre.
• Left arm OP = ℓ/4; right arm OQ = 3ℓ/4.
• B uniform, into the page; ω constant.

For a rod of length r pivoted at one end: ε = ½Bωr². Here the pivot is in the middle of the rod, so treat it as two separate rods sharing a common pivot O. Crucially, in both arms the force qv×B pushes charge outward, away from O. So O is one polarity and both ends are the other — the two emfs partially cancel when you go end-to-end.

1. Left arm: V_O − V_P = ½Bω(ℓ/4)² = ½Bωℓ²/16.
2. Right arm: V_O − V_Q = ½Bω(3ℓ/4)² = ½Bω·9ℓ²/16.
3. Subtract to eliminate V_O: V_P − V_Q = ½Bω[(3ℓ/4)² − (ℓ/4)²].
4. Bracket: 9ℓ²/16 − ℓ²/16 = 8ℓ²/16 = ℓ²/2.
5. ε = ½Bω·(ℓ²/2) = ¼Bωℓ².

• Difference-of-squares shortcut: ε = ½Bω(r₂² − r₁²) = ½Bω(r₂+r₁)(r₂−r₁). Here = ½Bω(ℓ)(ℓ/2) = ¼Bωℓ² — one line.
• Sanity anchors: pivot at the exact centre ⇒ difference = 0 ⇒ emf = zero. Pivot at one end ⇒ ½Bωℓ². Our pivot is between those cases, so the answer must lie strictly between 0 and ½Bωℓ² — which instantly kills both “Zero” and “½Bωℓ²”.

• B⃗ = î + 3k̂ — this lies entirely in the x–z plane (no ĵ component).
• v⃗ = 3î + 2ĵ + k̂
• Wire length 1 m, perpendicular to the plane of B ⇒ the wire points along ĵ, so L⃗ = 1ĵ.

Motional emf is a triple scalar product: ε = (v⃗ × B⃗)·L⃗. Do it in two clean stages — cross product first (gives a vector), then dot with the wire vector (gives a number). The key reading-comprehension step: “perpendicular to the plane of B” means perpendicular to the plane B lies in; since B has no ĵ part, B lives in the x–z plane and the wire is along ĵ.

1. Cross product, determinant form:
   v⃗×B⃗ = | î ĵ k̂ ; 3 2 1 ; 1 0 3 |
2. î component: (2)(3) − (1)(0) = 6.
3. ĵ component: −[(3)(3) − (1)(1)] = −(9 − 1) = −8.
4. k̂ component: (3)(0) − (2)(1) = −2.
5. So v⃗×B⃗ = 6î − 8ĵ − 2k̂.
6. Dot with L⃗ = 1ĵ: only the ĵ term survives ⇒ ε = −8 V.
7. Magnitude = 8 V.

• Don't compute the whole cross product. Since you will dot with ĵ anyway, you only need the ĵ component: −(v_xB_z − v_zB_x) = −(3·3 − 1·1) = −8. One line, no determinant.
• Watch the middle sign. The ĵ slot of a determinant carries a built-in minus. Dropping it is what turns 8 into the distractor answers.
• Note that v_y = 2 never appears — motion along the wire never produces emf. Useful sanity check.

• Long straight wire, current upward, increasing.
• Ring A to the left, ring B to the right, both in the plane of the page.

Two steps, in this order — never skip step 1:

1. Right-hand grip: point the right thumb up along the current. The fingers curl out of the page on the left and into the page on the right. So the two rings sit in opposite field directions — that alone tells you the answers must differ.
2. Lenz: the induced current opposes the change. Field increasing out of page ⇒ induced current makes field into page ⇒ clockwise. Field increasing into page ⇒ induced current makes field out of page ⇒ anticlockwise.

1. Ring A (left): B is out of the page and growing.
2. Oppose it ⇒ induced field must point into the page inside A.
3. Right-hand rule for a loop: into-page field ⇒ current runs clockwise.
4. Ring B (right): B is into the page and growing.
5. Oppose it ⇒ induced field must point out of the page inside B ⇒ current runs anticlockwise.

• Elimination first: because the field is out-of-page on one side and into-page on the other, the two rings cannot match. Options 3 and 4 die instantly — a 50/50 in two seconds.
• Clock mnemonic: “Flux out and growing → Clockwise” (out–grow–clock). Reverse either condition and you flip the sense.
• “Slowly increased” only guarantees the current is small; the direction depends purely on the sign of dΦ/dt, which is positive here.

• Long wire with current i pointing right (+x), rod AB vertical below it.
• Rod velocity v to the right (+x), parallel to the wire.
• A = top end (closer to wire), B = bottom end.

Even though the rod moves parallel to the wire (so the distance from the wire never changes and no flux sweeps), each free electron in the rod still feels F⃗ = qv⃗×B⃗. That force pushes charge along the rod and sets up a genuine potential difference. Motional emf does not require the field to be uniform or the flux to change — it only requires v⃗×B⃗ to have a component along the rod.

1. Field at the rod. Current is +î; the rod is below the wire, so the position vector from wire to rod is −ĵ. Grip rule ⇒ B points into the page: B⃗ = −Bk̂.
2. Force on a positive carrier: F⃗ = q(vî) × (−Bk̂) = −qvB(î×k̂).
3. Use î×k̂ = −ĵ ⇒ F⃗ = +qvBĵ, i.e. upward.
4. Positive charge therefore piles up at the top end = A.
5. Hence V_A > V_B.

• Kill two options instantly: “no emf” and “same potential” say the same thing. Two options that are physically identical can't both be right — so both are wrong. That leaves a 50/50 between A-higher and A-lower.
• Flat-hand rule: right palm flat, fingers along v (right), palm facing the way B goes (into page) → thumb gives the push on positive charge = up = toward A.
• Don't be distracted by B being non-uniform (it falls off as 1/r). Non-uniformity affects the size of the emf, never the polarity.

• Two identical inductors, each of self-inductance L.
• P: a and b straddle only the upper coil, which carries the full current I.
• Q: a and b straddle both coils in parallel; the current splits.

An inductor's emf is E = L·(dI_{through that coil}/dt). So the whole question reduces to: how much current actually flows through the element(s) sitting between a and b? Read the figure for where a and b are tapped — not for how the coils look.

1. Configuration P. Point b is tapped between the two coils, so only the upper coil lies between a and b. It carries the whole current I.
2. E_P = L·dI/dt.
3. Configuration Q. The two identical coils are in parallel, so the current splits equally: each carries I/2.
4. Emf across either branch (and hence across a–b): E_Q = L·d(I/2)/dt = (L/2)·dI/dt.
5. Ratio: E_P/E_Q = L / (L/2) = 2.

• Equivalent-inductance shortcut for Q: two L's in parallel give L_eq = L/2. For P, only one coil is spanned, so L_eq = L. Ratio = 2.
• Where this question is won or lost: if you assume P is a full series pair you get 2L/(L/2) = 4 — not even an option, which is your signal that you misread the tap point. If your answer isn't listed, re-read the figure, don't re-do the algebra.
• Reversing a ratio flips 2 into 1/2 — both are offered. Always write the ratio in the order the question names it: P on top.

A metal plate heats up only when real current flows inside it (H = I²Rt). Current arises in exactly two ways: you inject it from outside, or an emf is induced. Faraday says induction needs dΦ/dt ≠ 0 — a change in time. A field that varies only from place to place but is frozen in time gives dΦ/dt = 0 for a stationary plate ⇒ no eddy currents ⇒ no heating.

1. (a) DC passing through the plate ⇒ I²R heating. TRUE.
2. (b) Time-varying B ⇒ dΦ/dt ≠ 0 ⇒ eddy currents ⇒ heating. TRUE.
3. (c) Spatially varying but time-independent B, plate at rest ⇒ dΦ/dt = 0 ⇒ no eddy currents. FALSE.
4. (d) Any current, AC or DC ⇒ Joule heating regardless of direction. TRUE.
5. So the true set is a, b, d.

• Nesting check: statement (d) is the general version of (a). If (a) is true then (d) must be true — so any option containing (a) but not (d), or vice versa, is self-inconsistent. This kills bad options fast.
• The time test: for induction, cover the word “varying” and ask “varying with what?” Only with time counts. “Varying in space” is a deliberate decoy that appears in this exact question every year.
• Since a and d are obviously true, the entire question is really the single yes/no: is (b) or (c) the inducing one? Answer: (b).

• Square loop, side l, resistance R, speed v to the right.
• Region 1: B₀ into the page. Region 2: 2B₀ out of the page.
• The loop straddles the boundary at the instant considered.

The two fields are antiparallel, so when you write the total flux they carry opposite signs — and their contributions to the emf therefore add rather than subtract. This is the entire trap of the question. Equivalently: the one wire that lies on the boundary is being asked to do the work of both field regions at once.

1. Let x = length of loop already inside region 2. Take into the page as positive.
2. Flux: Φ = B₀·l(l−x) − 2B₀·lx.
3. Simplify: Φ = B₀l² − 3B₀lx.
4. Differentiate, with dx/dt = v: dΦ/dt = −3B₀lv.
5. Magnitude: |ε| = 3B₀lv ⇒ i = 3B₀lv/R.
6. Direction: into-page flux is decreasing ⇒ the loop fights back by trying to keep flux into the page ⇒ induced current is clockwise.

• The “add the fields” shortcut: for a wire on a boundary between opposite fields, the effective field is B₁ + B₂ = B₀ + 2B₀ = 3B₀, so ε = 3B₀lv in one step. If the two fields were in the same direction you would subtract instead.
• The number 3 is your signature. Seeing a “3” in the options when the data says B₀ and 2B₀ is a strong hint that the opposite-direction addition is the intended route.
• Direction shortcut: ask “which flux am I losing?” You're losing into-page flux → current flows so as to make into-page flux → clockwise.

• Rail separation = l; rod length = L with L > l — the rod overhangs.
• Total circuit resistance R, mass m, rails smooth (frictionless), field B uniform.

Terminal means acceleration has died to zero, so magnetic retarding force exactly balances weight. The subtlety: current flows only through the part of the rod between the rails. The overhanging pieces carry no current and contribute nothing, so the effective length is l, not L.

1. emf generated: ε = Blv (uses l).
2. Current: i = Blv/R.
3. Retarding force on the current-carrying part: F = Bil = B²l²v/R.
4. Terminal condition: mg = B²l²v_T/R.
5. Solve: v_T = mgR / B²l².

• The universal terminal-velocity template: v_T = (driving force)×R / B²l². On a vertical loop the driving force is mg; on an incline it becomes mg sinθ. Memorise the shape and you only ever swap the numerator.
• Trap detector: the problem takes the trouble to say L > l. That sentence exists only to bait you into writing L. Any option with L² is a plant — strike it out on sight.
• No stray factors of 2 or 8 can appear — the force balance is a single clean equation. Options with 2 or 8 are pure noise.

• ε = 25 V, i₁ = 10 A, i₂ = 25 A, Δt = 1 s.
• “Uniform rate” ⇒ di/dt is constant.

Two formulas chained together:

The trick is that L is hidden: the problem hands you the emf so you can back out L first.

1. Rate of change: di/dt = (25 − 10)/1 = 15 A/s.
2. Find L: 25 = L × 15 ⇒ L = 25/15 = 5/3 H.
3. Energy change: ΔU = ½L(i₂² − i₁²) = ½(5/3)(625 − 100).
4. Bracket: 625 − 100 = 525.
5. ΔU = (5/6) × 525 = 2625/6 = 437.5 J.

• Keep L as a fraction. 5/3 stays exact; 1.667 introduces rounding and makes 437.5 look like 437.4 or 438. Fractions until the last line.
• Difference of squares: 25² − 10² = (25+10)(25−10) = 35 × 15 = 525 — faster and less error-prone than squaring both.
• Cancel early: ½ × (5/3) × 35 × 15 → the 15 cancels the 3 → ½ × 5 × 35 × 5 = 437.5.
• The distractor 637.5 J is what you get from ½L(i₂−i₁)²-style slips — energy goes with the difference of squares, never the square of the difference.

• Phase 1 (X–Y, long time): cell E drives a steady current through L and R₁.
• Phase 2 (X–Z): the cell is cut out; L is left in a closed loop with R₂.

Pure energy bookkeeping — no differential equations needed.

• Long time ⇒ inductor behaves as a plain wire ⇒ steady current i₀ = E/R₁, and the inductor has banked U = ½Li₀².
• After switching, R₂ is the only resistor in the decay loop, so every joule stored in L must end up as heat in R₂.

1. Steady current at the end of phase 1: i₀ = E/R₁.
2. Energy stored in L: U = ½Li₀² = ½L(E/R₁)².
3. Simplify: U = LE²/(2R₁²).
4. In phase 2 the current decays to zero through R₂ alone, so H_R2 = U.
5. H = LE²/(2R₁²).

• Dimension sieve: energy must not depend on R₂ at all — R₂ only controls how fast the heat comes out, not how much. That single insight eliminates three of the four options immediately.
• Slow/fast irrelevance: halving R₂ doubles the decay time and halves the power — the product (total energy) is unchanged.
• “Long time” = inductor is a wire. Pair this with the Q6 rule (“just closed” = inductor is a break) and you own both switching extremes.

• Linear density λ = m/L = 0.45 kg/m.
• θ = 45°, B = 0.15 T vertical, g = 10 m/s².
• Incline is smooth ⇒ no friction; only three forces act.

The rod is horizontal and the field is vertical, so F⃗ = I L⃗×B⃗ is horizontal (perpendicular to both), pushing the rod into the hill. Resolve along the incline surface so the normal force drops out:

• Weight component pulling it down the slope: mg sinθ.
• Magnetic force component pushing it up the slope: F cosθ (F is horizontal, so its along-slope part carries cosθ).

1. Magnetic force magnitude: F = BIL, directed horizontally.
2. Balance along the incline: BIL cosθ = mg sinθ.
3. Divide by cosθ: BIL = mg tanθ.
4. Divide by L to bring in linear density: BI = (m/L)g tanθ = λg tanθ.
5. Insert numbers, with tan45° = 1: I = (0.45 × 10 × 1)/0.15.
6. I = 4.5/0.15 = 30 A.

• Memorise the compact form: I = λg tanθ / B. It works for every version of this question; only θ changes.
• tan45° = 1 means the angle contributes nothing here — the whole problem is 0.45×10/0.15. Don't build a full free-body diagram for an arithmetic question.
• Why cos and not sin: the magnetic force is horizontal, the weight is vertical. Whichever force is along the direction you'd expect gets sin; the perpendicular one gets cos. Sketch the two arrows before resolving — that one sketch prevents the commonest swap.
• Fast check: 0.45/0.15 = 3, times 10 = 30. Done mentally.

• Disk carries positive charge, spinning clockwise in the plane of the page.
• Point A is in the same plane, outside the disk's rim.

Three linked ideas:

• Spinning positive charge = conventional current in the same sense as the rotation ⇒ clockwise current loop. (Had the charge been negative, the current would be anticlockwise and every answer flips.)
• A clockwise loop produces B into the page inside the loop.
• Field lines are closed loops: what goes down through the middle must come back up outside. So at any external in-plane point the field points out of the page.

1. Positive charge + clockwise spin ⇒ conventional current is clockwise.
2. Right-hand rule for the loop: curl fingers clockwise ⇒ thumb points into the page ⇒ B is into the page inside the disk.
3. Magnetic field lines form closed loops (∇·B = 0; there are no magnetic charges).
4. Therefore the return path lies outside the rim, pointing the opposite way: out of the page at A.
5. Since the return field there is perpendicular to the disk plane, the answer is “perpendicular, out of the page”.

• Bar-magnet analogy: a spinning charged disk is a flat magnet. Field exits the N face, loops round the outside, re-enters the S face. Anywhere beside a magnet, the external field opposes the internal field — exactly our situation.
• The angle options are decoys. By symmetry (the disk is symmetric about its own plane), the field at an in-plane point must be perpendicular to that plane — 30° and 60° are impossible. That is a 50/50 for free.
• Then the only question is in or out, and “outside is always opposite to inside” settles it.

• Initial radius r = 4 cm.
• Final kinetic energy K′ = K/2.
• B and q unchanged (same field, same particle).

r = mv/(qB) = p/(qB), so r ∝ p (momentum), not energy. And p = √(2mK), so r ∝ √K. Energy enters under a square root — halving K does not halve r.

1. Write the proportionality: r ∝ √K.
2. Form the ratio: r′/r = √(K′/K) = √(1/2) = 1/√2.
3. Substitute: r′ = 4/√2.
4. Rationalise: 4/√2 = 4√2/2 = 2√2.
5. r′ = 2√2 cm ≈ 2.83 cm.

• Direction check before arithmetic: the particle lost energy, so it must move slower, so the circle must get smaller. Any answer ≥ 4 cm is impossible. That kills “4 cm” and “8 cm” before you calculate anything.
• The √2 signature: whenever a question halves or doubles an energy and asks for r, p, or v, expect a √2 in the answer. Seeing 2√2 among plain integers is the examiner's tell.
• Useful chain to memorise: r ∝ p ∝ v ∝ √K. And for the same K but different charge: r ∝ √m / q.

• Flexible wire of total length l forming a closed loop.
• B normal (perpendicular) to the loop plane; current I.

Two stages:

• Magnetic forces on a flexible current loop push it outward everywhere, so it settles into the shape with the largest area for a given perimeter — a circle. Radius from perimeter: l = 2πR ⇒ R = l/2π.
• Take a tiny arc subtending dθ. The outward magnetic force on it is balanced by the inward resultant of the two tensions at its ends — exactly the same geometry as a spinning ring or a belt over a pulley.

1. Element length: dl = R dθ. Outward magnetic force: dF = BI·R dθ.
2. The two tension vectors at the element's ends each tilt inward by dθ/2; their inward resultant is 2T sin(dθ/2).
3. Small-angle approximation: sin(dθ/2) ≈ dθ/2, so the inward pull is 2T·(dθ/2) = T dθ.
4. Equilibrium: BIR dθ = T dθ ⇒ the dθ cancels ⇒ T = BIR.
5. Substitute R = l/2π: T = BIl/2π.

• Remember the master result: T = BIR for any circular current loop in a normal field. Then just express R in whatever the question gives you (here, the perimeter).
• Spot the π. The perimeter was given, not the radius — so a π must appear in the answer. Options without π (½BIl and BIl) are therefore impossible, and (π/2)BIl has π on the wrong side. Pure structure, no derivation.
• Physical sanity: bigger loop → more tension; stronger field or current → more tension. T = BIl/2π satisfies all three.

• R = 0.2 m ⇒ A = πR² = 0.04π m².
• I = 1.0 A, clockwise when viewed from +z.
• B = 2 T at 45° in the x–y plane.

τ⃗ = m⃗×B⃗ where m⃗ = IA n̂. Two sign-sensitive facts: (i) clockwise seen from +z means n̂ = −k̂, and (ii) the loop lies in the x–y plane while B also lies in the x–y plane, so m⃗ is perpendicular to B⃗ and the torque is maximal.

1. Resolve B: B⃗ = 2cos45°î + 2sin45°ĵ = √2(î + ĵ) T.
2. Magnetic moment: m⃗ = IA n̂ = (1)(0.04π)(−k̂) = −0.04π k̂.
3. Cross product: τ⃗ = (−0.04π k̂) × √2(î + ĵ).
4. Use k̂×î = ĵ and k̂×ĵ = −î: τ⃗ = −0.04√2π(ĵ − î) = 0.04√2π(î − ĵ).
5. Numerically: 0.04 × 1.414 × 3.1416 = 0.1777 ≈ 0.18.
6. τ⃗ = 0.18(î − ĵ) N·m.

• Magnitude first, direction second. |τ| = mB = 0.04π × 2 = 0.251… and the answer vector 0.18(î−ĵ) has magnitude 0.18√2 = 0.254 ✓. Use this to confirm you haven't lost a factor.
• Only the sign is in dispute: options 1 and 3 are the same vector with opposite signs, so the entire question is “is n̂ along +k̂ or −k̂?” Clockwise from +z ⇒ −k̂. Write that down before the cross product.
• τ ⊥ B always. Check: (î−ĵ)·(î+ĵ) = 1 − 1 = 0 ✓. A free, instant verification of the direction.

• Initially the whole main current I₀ goes through G.
• Shunt 1: S₁ = 4 Ω ⇒ galvanometer current falls to I₀/5.
• Then a 2 Ω wire is added as a further shunt (so 4 Ω and 2 Ω are both in parallel with G).
• Main current stays I₀ throughout.

Current divider between G and the shunt S:

Step 1 uses the “one-fifth” information to find the unknown G. Step 2 then reuses the same divider with the combined shunt.

1. First shunting: I₀/5 = I₀ × 4/(4 + G).
2. Cancel I₀ and cross-multiply: 4 + G = 20 ⇒ G = 16 Ω.
3. Combined shunt (4 Ω parallel 2 Ω): S = (4×2)/(4+2) = 8/6 = 4/3 Ω.
4. Apply the divider again: I_g = I₀ × (4/3) / ((4/3) + 16).
5. Denominator: 4/3 + 48/3 = 52/3.
6. I_g = I₀(4/3)÷(52/3) = 4I₀/52 = I₀/13.

• “Deflection reduced to 1/n” gives G in one line: G = (n − 1)S. Here G = (5−1)×4 = 16 Ω — no algebra at all.
• Reuse it in reverse for the answer: with combined shunt S = 4/3, the reduction factor is n = 1 + G/S = 1 + 16÷(4/3) = 1 + 12 = 13, so I_g = I₀/13. Two lines total.
• Read “further shunted” carefully: it means added in parallel to what's already there, not replaced by. Using 2 Ω alone gives I₀/9 — not an option, which is your warning flag.

• m = 50 g = 0.05 kg, L = 10 cm = 0.1 m.
• B = 5 mT = 5×10⁻³ T, I = 2 A (constant, maintained by the source).
• u = 0, t = 10 s, frictionless.

Because the source forces the current to stay at 2 A, the back-emf never reduces it. So F = BIL is constant ⇒ acceleration is constant ⇒ plain kinematics v = u + at applies. (Contrast with Q20, where the current was induced, grew with speed, and produced a terminal velocity instead.)

1. Force: F = BIL = (5×10⁻³)(2)(0.1) = 10⁻³ N.
2. Acceleration: a = F/m = 10⁻³/0.05.
3. a = 0.02 m/s².
4. Kinematics: v = 0 + (0.02)(10) = 0.2 m/s.
5. Convert: 0.2 m/s = 20 cm/s.

• One-line combined formula: v = BILt/m. Plug once: (5×10⁻³×2×0.1×10)/0.05 = 0.2.
• Unit discipline is the whole risk here: g→kg, cm→m, mT→T. Convert all three before any multiplication, and note the answer is requested in cm/s, so convert back at the end. Three of the four distractors are unit-slip landing zones.
• Sanity: a is tiny (0.02 m/s²), so after 10 s the speed should be small — a few tens of cm/s. 20 cm/s fits.

• I = 10 A, B = 5 T out of the page.
• Path from P to Q: 4 cm up, 6 cm across, 4 cm up.

Effective length theorem: for any wire (however bent) carrying current I in a uniform field, the net force is

where L⃗_eff is the straight-line vector from start to finish. All the intermediate zig-zags cancel. (For a closed loop the start equals the finish, so L_eff = 0 and the net force is zero — but this wire is open.)

1. Displacement components P→Q: horizontal = 6 cm, vertical = 4 + 4 = 8 cm.
2. Effective length: L_eff = √(6² + 8²) = √100 = 10 cm = 0.1 m.
3. The 6–8–10 triangle — a standard Pythagorean triple.
4. L⃗_eff lies in the page and B is perpendicular to the page, so the angle is 90° and |L⃗×B⃗| = L_effB.
5. F = BIL_eff = 5 × 10 × 0.1 = 5 N.

• Never add segment lengths. Adding 4+6+4 = 14 cm gives 7 N — and 7 N isn't even offered, which is the examiner telling you that route is wrong. Add the vectors, not the lengths.
• Look for Pythagorean triples in these figures: 3–4–5, 6–8–10, 5–12–13. The numbers are chosen so L_eff is a whole number.
• Open vs closed test: before anything else ask “do the ends meet?” Closed → F = 0. Open → join the ends with a straight line and use that.

• Arc length L = 10 cm = 0.1 m, angle θ = 1 rad, current I = 1 A.
• μ₀/4π = 10⁻⁷ T·m/A.

For an arc subtending angle θ (in radians) at radius r:

The radius is not given directly — it comes from the arc-length relation L = rθ. Since θ = 1 rad, r = L = 0.1 m, which is what makes this question quick.

1. Find r: L = rθ ⇒ 0.1 = r × 1 ⇒ r = 0.1 m.
2. Apply the arc formula: B = 10⁻⁷ × (1 × 1)/0.1.
3. Divide: 10⁻⁷/10⁻¹ = 10⁻⁶.
4. B = 1 × 10⁻⁶ T.

• Rewrite the formula in terms of arc length: since θ = L/r, we get B = (μ₀/4π)·IL/r². Sometimes that's the faster route.
• Fraction-of-a-circle view: a full loop gives B = μ₀I/2r; an arc of θ radians gives θ/2π of that. Check: (1/2π)(4π×10⁻⁷×1)/(2×0.1) = 10⁻⁶ ✓.
• Degrees vs radians: the formula only works in radians. If a question says 60°, convert to π/3 first. Here “1 radian” is handed to you — a gift, not a complication.

S1 analysis. Around a long straight wire, B forms circles concentric with the wire. For the particle to travel a circle around the wire, its velocity would have to be tangential — i.e. parallel to B. But F = qv×B = 0 when v ∥ B. With no centripetal force available, that orbit cannot be sustained. S1 is TRUE.

S2 analysis. For any particle actually moving in a circle in a magnetic field, the magnetic force is the centripetal force — it points toward the centre, within the plane of the circle, not normal to it. So S2 is a false general claim. S2 is FALSE. (It also says “electromagnetic”, which would include electric forces, and those can point any way at all.)

1. Judge S1 alone. Circular orbit around the wire ⇒ v is parallel to B ⇒ F = qv×B = 0 ⇒ no centripetal force ⇒ impossible. TRUE.
2. Judge S2 alone, as a general statement. Standard circular motion in a field (e.g. B out of the page, v in the page): the force qv×B lies in the plane, pointing at the centre. FALSE.
3. Combine. True + False ⇒ only one option survives.
4. No explanation question arises, since S2 is false.

• The fixed three-step defence (use it for every assertion–reason item, without exception):
    1. Write T/F for S1, ignoring S2 entirely.
    2. Write T/F for S2 as a standalone general claim, ignoring S1.
    3. Only if both are T, ask about explanation.
  Most A–R errors come from steps 1 and 2 contaminating each other.
• Absolute words are usually false. S2 says the force is normal to the plane — a universal claim about all circular motion. One counterexample (any ordinary cyclotron orbit) destroys it.
• If S2 were true, magnetic circular motion could not exist at all — yet cyclotrons work. That contradiction is the fastest route to “S2 false”.

• Solid cylinder radius a, current i uniform across the cross-section.
• One field point inside at distance r; one outside at distance R; the two field magnitudes are equal.

Ampere's law ∮B·dl = μ₀I_enclosed gives two different behaviours:

Inside, only the fraction i(r²/a²) is enclosed; outside, the whole current is enclosed.

1. Inside: enclosed current = i·(πr²)/(πa²) = i r²/a²; Ampere gives B·2πr = μ₀ir²/a² ⇒ B_in = μ₀ir/(2πa²).
2. Outside: B·2πR = μ₀i ⇒ B_out = μ₀i/(2πR).
3. Set them equal: μ₀ir/(2πa²) = μ₀i/(2πR).
4. Cancel μ₀i/2π from both sides: r/a² = 1/R.
5. Cross-multiply: a² = Rr ⇒ a = √(Rr).

• Geometric-mean signature: whenever a ∝-r law is set equal to a ∝-1/r law, the crossing point is always a geometric mean. Recognising √(Rr) as “the GM answer” lets you pick it on sight.
• Dimensional check: a, r, R are all lengths, so any correct expression must be degree 1 in length. √(Rr) ✓. R²/r ✓ dimensionally but fails the limit test below; Rr/(R+r) ✓ too — so use the next check to separate them.
• Limit test: if r → a then the point is on the surface and by continuity R must also → a. Only a = √(Rr) satisfies that cleanly.

• Slab width d = mv/(√2 qB).
• Particle enters perpendicular to the boundary with speed v; field B into the page.
• Radius of the circular path: r = mv/(qB).

Inside the slab the particle follows a circular arc of radius r = mv/(qB). If it enters perpendicular to the boundary, the penetration depth d and the deviation angle θ are linked by

Then the time is just the fraction of a full period: t = (θ/2π)T with T = 2πm/(qB), which simplifies beautifully to t = θ/ω = θm/(qB).

1. Radius: r = mv/(qB).
2. Ratio: sinθ = d/r = [mv/(√2 qB)] ÷ [mv/(qB)] = 1/√2.
3. So θ = 45° = π/4 rad.
4. Cyclotron angular frequency: ω = qB/m.
5. Time: t = θ/ω = (π/4) ÷ (qB/m).
6. t = πm/(4qB).

• Learn the two-line template: sinθ = d/r, then t = θm/(qB). Every “time in a field slab” question is these two lines.
• The width is written to reveal the answer. Seeing mv/(√2 qB) — i.e. r/√2 — should trigger “sinθ = 1/√2, so 45°” before you write anything.
• Structure check: the answer must contain a π (it's a fraction of a circle) and must not contain v (the period of circular motion is speed-independent). That alone eliminates every distractor except πm/4qB and πm/2qB, and 45° ≠ 90° settles it.

• Long straight wire carrying i₁; rectangular loop carrying i₂.
• Both lie in the same plane; near side at distance a, far side at distance b.

Torque on a magnetic dipole is τ⃗ = m⃗×B⃗, which vanishes when m⃗ ∥ B⃗ (or antiparallel). Here the loop lies in the plane of the page, so m⃗ is perpendicular to the page. The field from the straight wire at the loop's location is also perpendicular to the page (into it, on that side). So m⃗ and B⃗ are along the same line ⇒ τ = 0.

Force-level view: the two sides parallel to the wire feel unequal forces (attraction at a, repulsion at b), giving a net force. The two perpendicular sides feel equal and opposite forces along the same line. No pair produces a couple — so no torque, even though there is a net force.

1. Loop plane = page ⇒ m⃗ = ±(i₂A)k̂, perpendicular to the page.
2. Field from the straight wire at every point of the loop is also perpendicular to the page (same side of the wire ⇒ same direction).
3. Hence m⃗ and B⃗ are collinear.
4. τ = mB sin0° = 0.
5. Cross-check at force level: forces on the near and far sides are antiparallel and act along the same line of action, so they produce no couple.

• Coplanar test: if a current loop and a straight wire lie in the same plane, the torque about the loop's centre is always zero. Only the net force survives. Memorise this as a one-line rule.
• “Zero” is a real answer. Being offered an elaborate algebraic expression alongside “zero” is a classic examiner set-up; the ugly options are constructed from the force formula to bait you.
• Force vs torque audit: underline the asked quantity first. The expression (μ₀/2π)i₁i₂l(b−a)/ab is precisely the net force — a correct formula answering the wrong question.

• Two loops, radius a, current I, planes mutually perpendicular.
• Point C is the centre of one loop and lies on the axis of the other at axial distance x = a.

Two standard results, then vector addition — the fields are perpendicular to each other because the loops are perpendicular:

1. Loop 1 (C is its centre). B₁ = μ₀I/2a, directed along that loop's axis — take it as +k̂.
2. Loop 2 (C is on its axis, x = a). Denominator: (a²+a²)^3/2 = (2a²)^3/2 = 2√2 a³.
3. B₂ = μ₀Ia²/(2 × 2√2 a³) = μ₀I/(4√2 a), directed along loop 2's axis — take it as −î.
4. Express both over the common factor μ₀I/(4√2 a): note B₁ = μ₀I/2a = (μ₀I/4√2 a) × 2√2.
5. Add: B⃗ = (μ₀I/4√2 a)(−î + 2√2 k̂).
6. Factor out the minus: B⃗ = −(μ₀I/4√2 a)(î − 2√2 k̂).

• Ratio shortcut: B_centre/B_axial = (μ₀I/2a) ÷ (μ₀I/4√2a) = 2√2. So the centre-field component must be 2√2 times bigger. Just find which option has the larger coefficient on the correct axis — that alone picks the answer.
• Memorise the x = a case: on the axis at a distance equal to the radius, the field is μ₀I/(4√2 a), i.e. 1/(2√2) of the centre value. It recurs constantly.
• Perpendicular loops ⇒ perpendicular fields ⇒ components never mix. No cosθ factors appear anywhere.

The law is F/L = μ₀i₁i₂/(2πR) — an inverse relationship. Plotting F/L against R gives a rectangular hyperbola, not a straight line. Note the irony: Statement 2 is exactly the evidence that Statement 1 is false.

1. S2 first (it's the easier one). The formula μ₀i²/2πR is correct, and like currents do attract. S2 TRUE.
2. Now S1. Using S2's own formula, F/L ∝ 1/R.
3. A y = k/x relation graphs as a hyperbola. A straight line would require F/L ∝ R. S1 FALSE.
4. Combine: False + True ⇒ option 4.
5. (A plot of F/L versus 1/R would be a straight line — that is the confusion the question is built on.)

• Do the easier statement first. Here S2 hands you the formula that decides S1. Order of attack matters in assertion–reason questions.
• “Straight line” keyword audit: a straight-line graph requires y = mx + c. Ask “is the relation linear in the plotted variable?” Inverse-square and inverse laws never plot straight against the raw variable.
• Compare with Q34 in this same set, where the pattern was True + False. Assertion–reason questions in this paper broke both ways — more evidence that the procedure, not the content, needs fixing.

• AB lies along the x-axis carrying I; PQ lies along the y-axis carrying 4 A.
• Null point at S(4, 3) — so its perpendicular distance from AB (the x-axis) is 3, and from PQ (the y-axis) is 4.

“Null point” means the two fields cancel there: equal magnitudes, opposite directions. Since the wires are perpendicular, at S both fields are perpendicular to the page (one out, one in), so cancellation is possible and reduces to a magnitude equation:

The key comprehension step: the distance from a point to a wire is the perpendicular distance, so the roles of 4 and 3 are crossed relative to the coordinates.

1. Perpendicular distance of S from AB (the x-axis) = the y-coordinate = 3.
2. Perpendicular distance of S from PQ (the y-axis) = the x-coordinate = 4.
3. Set the magnitudes equal: I/3 = 4/4.
4. Right-hand side: 4/4 = 1.
5. I = 3 A.

• The cross-over check: distance from the x-axis is y, distance from the y-axis is x. Writing I/4 = 4/3 instead gives 16/3 ≈ 5.3 — close to the decoy “5 A”, which is exactly why that option is there.
• The μ₀/2π always cancels in null-point problems. Jump straight to I₁/d₁ = I₂/d₂.
• Direction feasibility: a null point can only exist where the two fields oppose. S(4,3) is in the first quadrant — verify with the grip rule that one field is out of the page and the other into it. If they pointed the same way, no null point could exist anywhere in that quadrant.

• Radius R; two quarter arcs AB and CD, each 90°.
• Four long straight sections: A′A and BB′ (one current path), C′C and DD′ (the other).

Handle the pieces with three standard results:

• Semi-infinite wire ending at the foot of the perpendicular from O: B = μ₀i/(4πR).
• Wire whose line passes through O: B = 0 (every element has dl⃗ ∥ r⃗, so dl⃗×r̂ = 0).
• Quarter arc: B = μ₀i/(8R).

1. Wires C′C and DD′ run along lines through O, so each contributes exactly zero.
2. Wires A′A and BB′ are semi-infinite at perpendicular distance R, each contributing μ₀i/(4πR).
3. Check their directions with the grip rule: both come out of the page ⇒ they add: 2 × μ₀i/(4πR) = μ₀i/(2πR).
4. Arc AB is traversed anticlockwise ⇒ contributes μ₀i/(8R) out of the page. Arc CD is traversed clockwise ⇒ μ₀i/(8R) into the page.
5. The two arcs are equal and opposite ⇒ they cancel exactly.
6. Net: B = μ₀i/(2πR), out of the page.

• Scan for zero-contributors first. Any straight wire whose line passes through the field point contributes nothing. Striking those out early removes most of the work.
• The π audit. Arcs bring a bare 1/R with no π in the denominator; straight wires bring 1/πR. The correct option has π in the denominator and no loose π in the numerator — which is exactly what you'd expect if the arcs cancelled. Option 4, with its (π+1), is the “arcs added” trap.
• Symmetry sniff test: two equal arcs traversed in opposite senses almost always cancel. Check the current arrows before computing either one.

• Same wire length L used for both coils.
• N_A = 2, N_B = 4; same current i.

M = NiA = Niπr². The catch: more turns from the same wire means a smaller radius, and the radius enters squared — so the shrinking radius beats the growing turn count.

From L = N(2πr) we get r = L/(2πN), hence

1. Coil A: L = 2(2πr_A) ⇒ r_A = L/4π.
2. M_A = 2iπr_A² = 2iπL²/16π² = iL²/8π.
3. Coil B: L = 4(2πr_B) ⇒ r_B = L/8π.
4. M_B = 4iπr_B² = 4iπL²/64π² = iL²/16π.
5. Ratio: M_A/M_B = (1/8π)÷(1/16π) = 2.

• Skip all algebra with M ∝ 1/N: M_A/M_B = N_B/N_A = 4/2 = 2. One line, if you've stored that proportionality.
• Physical sanity: the coil with fewer turns is bigger, and area beats turn count because area goes as r². So fewer turns → larger moment. That immediately rules out any answer below 1.
• Ratio-order discipline: the question says M_A/M_B. Write “A on top” before substituting. Both 2 and 1/2 are offered precisely to catch the reversal.

• AB horizontal carrying i₁; CD vertical carrying i₂; the two are perpendicular and coplanar.
• AB is free to move.

The field from CD falls off as 1/r, so along the length of AB the field is stronger at the end nearer CD (A) and weaker at the far end (B). Each element of AB feels dF = i₁(dl⃗×B⃗), and because B varies along AB:

• the forces do not cancel ⇒ there is a net force ⇒ translation;
• the forces are unequally distributed along the rod ⇒ a net torque about the centre of mass ⇒ rotation.

1. Field from CD at a point of AB at distance r: B = μ₀i₂/(2πr), directed perpendicular to the page.
2. Force per element: dF = i₁B dl = μ₀i₁i₂dl/(2πr), all in the same sense along AB.
3. Because every element's force points the same way, they cannot cancel ⇒ F_net ≠ 0 ⇒ the wire translates.
4. The near end A experiences a larger force than the far end B (1/r weighting).
5. Unequal forces at different distances from the centre of mass ⇒ net torque ⇒ the wire also rotates.
6. Hence both motions occur simultaneously.

• Two-question test for any “what motion?” item:
    (i) Do the forces cancel? No → translation.
    (ii) Are they uniformly distributed? No → rotation.
  Both answers are “no” here, so the answer is “both”.
• Non-uniform field → expect rotation. Only in a genuinely uniform field can you get pure translation.
• Contrast with parallel wires: two parallel wires sit at a constant separation, so B is the same along the whole length → pure translation (attraction or repulsion), no rotation. The perpendicular arrangement is what breaks that symmetry.

1. Circle the asked quantity and any negative word (NOT, INCORRECT, EXCEPT, interval, torque) before looking at the options.
2. For every ratio, write the empty fraction with the named term already on top, then fill it in.
3. For assertion–reason, write T/F for statement 1 alone, then T/F for statement 2 alone as a general claim, and only then consider the explanation link.